Solving Spring-Carts Exploding: Find Speed of M1 & M2

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A massless spring with a spring constant of 20 N/m compresses between two carts, M1 (5 kg) and M2 (3.5 kg), by 1.8 m before being released. The conservation of energy and momentum principles apply, leading to the equations for initial potential energy equating to final kinetic energy. The final potential energy of the spring is zero once released. To solve for the speeds of the carts after release, two equations are needed: one from energy conservation and another from momentum conservation. This approach allows for solving the two unknowns, the final velocities of the carts.
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A massless spring of spring constant 20 N/m is placed between two carts. Cart 1 has a mass M1 = 5 kg and Cart 2 has a mass M2 = 3.5 kg. The carts are pushed toward one another until the spring is compressed a distance 1.8 m. The carts are then released and the spring pushes them apart. After the carts are free of the spring, what are their speeds?

i know both momentum and energy conservation applies here, but don't know where to start...
both carts are initially at rest,

so I'm guessing : PE(initial) = PE(final) + KE (final)
 
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You are correct: Conservation of energy and momentum is the way to deal with this problem. So your starting point should be obvious: Write down the initial energy and momentum (these are known) and equate them to the final energy and momentum respectively. This will give you 2 equations in 2 unknowns (v1 and v2).
 
also, wouldn't PE final be zero too? after the carts are released, the spring would be relaxed again, thus no compression, thus zero..

that'll make it: initial PE = final KE

but i don't know where to go from here
 
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That's right, the final potential energy of the spring is zero.
What are the expressions for the initial energy, initial momentum, final momentum and final energy?
 
now we have initial PE = final KE,

it goes...

.5kx^2 = .5m(1)v(1)^2 + .5m(2)v(2)^2

does it make sense? but the problem now is i have two variables to solve...one equation...
 
Don't forget conservation of momentum. That will give you the second equation that you need.
 
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